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What I am going to ask is probably simple and maybe trivial. But I want to be sure that I am not missing any point. Let ${\bf G}\subseteq \mathrm{GL}_n(\mathbb{C})$ be a simple and simply connected group. Then this group is defined over $\mathbb{Z}$. Hence I can talk about ${\bf G}(\mathbb{F}_q)$ where the characteristic of the finite field $\mathbb{F}_q$ is $p$.

  1. Is it true that when $p\geq 3$, the finite group ${\bf G}(\mathbb{F}_q)$ is perfect? Of course since $\bf G$ is simple then $\bf G=\bf G'$ but does this imply that the commutator subgroup of ${\bf G}(\mathbb{F}_q)$ is the same as ${\bf G}(\mathbb{F}_q)$.
  2. Moreover is it true that ${\bf G}(\mathbb{F}_q)/(Z({\bf G}(\mathbb{F}_q)))$ is a simple group?

Thanks

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  • $\begingroup$ Essentially yes. See Theorem 24.17 in the book "Linear Algebraic Groups and Finite Groups of Lie Type" by Malle and Testerman. There are some exceptions for very small groups, generally when $q = 2$, but it's a small list. $\endgroup$
    – Jay Taylor
    Commented Oct 30, 2015 at 9:27
  • $\begingroup$ The questions aren't at research-level, so should be asked first at a site such as math.stackexchange. Also, the formulation isn't quite right: being defined over $\mathbb{Z}$ (whatever that means) isn't the point here: $G$ should be assumed to be defined and split over a finite prime field. (Here the assumption that $G$ is "simple" just means that it is simple in the sense of algebraic groups, not abstract groups.) The results Jay quotes go back to Chevalley, Steinberg, et al., and are found in many sources such as Steinberg's 1967-68 Yale lectures: math.ucla.edu/~rst $\endgroup$
    – Jim Humphreys
    Commented Oct 30, 2015 at 13:55

1 Answer 1

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I'll just make my comment an answer so as to close this question. By Theorem 24.17 of the book "Linear Algebraic Groups and Finite Groups of Lie Type" by Malle and Testerman we have the answer is yes unless $\mathbf{G}(\mathbb{F}_q)$ is one of the groups

$\mathrm{SL}_2(\mathbb{F}_2)$, $\mathrm{SL}_2(\mathbb{F}_3)$, $\mathrm{SU}_3(\mathbb{F}_2)$, $\mathrm{Sp}_4(\mathbb{F}_2)$, $\mathrm{G}_2(\mathbb{F}_2)$, ${}^2\mathrm{B}_2(\mathbb{F}_2)$, ${}^2\mathrm{G}_2(\mathbb{F}_3)$, ${}^2\mathrm{F}_4(\mathbb{F}_2)$.

These groups are genuine exceptions as is explained in Remark 24.18.

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  • $\begingroup$ As my comment on the question indicates, I think it should migrate to a more elementary forum since the results are old and well-known. $\endgroup$
    – Jim Humphreys
    Commented Oct 30, 2015 at 13:57
  • $\begingroup$ I agree. I'm happy for the question to be closed or migrated (the formulation by the author is also strange). Just thought it would be good to give some reference and finish the question. $\endgroup$
    – Jay Taylor
    Commented Oct 30, 2015 at 14:15
  • $\begingroup$ Dear Jay and Jim: Thanks for your comments. $\endgroup$
    – M.B
    Commented Oct 30, 2015 at 15:29

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